Yet Another Delta-Epsilon

  • #1

Homework Statement



Prove that lim (1/x) x --> 0 does not exist, i.e., show that lim (1/x) x --> 0 = l is false for every number l.

Homework Equations


0 < |x-a| < d
|f(x) - l| < E



The Attempt at a Solution



The strange thing is, the first time through I got the same solution as Spivak, but looking over it again the logic seems downright wrong. Here's the solution, verbatim:

The function f(x) = 1/x cannot approach a limit at 0, since it becomes arbitrarily large near 0. In fact, no matter what d > 0 may be, there is some x satisfying 0 < |x| < d, but 1/x > |l| + E, namely, any x < min(d, 1/(|l| + E)). Any such x does not satisfy |(1/x) - l| < E.

Where is he getting the bold portion from?

I write |1/x - l| < E
|1/x| - |l| < |1/x - l| < E
|1/x| - |l| < E
|1/x| < E + |l|


How is he getting greater-than?
 

Answers and Replies

  • #2
Office_Shredder
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I write |1/x - l| < E


Assuming this is true, then you're golden. But it's not going to be true for really small x, that's the point. So we see how he's doing a proof by contradiction: If the limit exists, you can't possibly get a greater than sign, but he gets a greater than sign. The way he gets it is by saying when x < min(d, 1/(|l| + E)), you get 1/x > |l| + E
 
  • #3
vela
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Where is he getting the bold portion from?
You're misreading what he wrote. He is merely asserting that there is an x such that 0<x<d and 1/x>|l|+E. Then he tells you how to find such an x.
I write |1/x - l| < E
|1/x| - |l| < |1/x - l| < E
|1/x| - |l| < E
|1/x| < E + |l|


How is he getting greater-than?
If the inequality you started with is true, then your conclusion is also true. If you go one step further and take the reciprocal of both sides, you get

[tex]|x| > \frac{1}{|l|+\epsilon}[/tex]

Spivak is saying if you choose an x such that this last inequality is false, the one you started with will also be false.
 
  • #4
So, he's essentially saying:

|1/x| < E + |l|

and then pointing out that, if d = 1/(|l| + E), this leads to a contradiction of that inequality, and therefore the whole thing violates the definition of a limit?

Oy! It's been almost a week and I'm still having trouble with these things.
 
  • #5
Office_Shredder
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So, he's essentially saying:

|1/x| < E + |l|

He isn't saying that because it's not going to be true. But it's what you should keep in the back of your mind... IF l was the limit, you would get this inequality. But in fact we get an opposite inequality for small x. Therefore, l is not the limit. Since l was arbitrary, there is no limit
 
  • #6
vela
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No, he's stating if 0<|x|<min{δ, 1/(|L|+ε)}, then 0<|x|<δ and |1/x|>|L|+ε. The δ appears only because the definition of the limit requires that |x|<δ. So long as |x| is small enough, you can satisfy both conditions. The point is, regardless of what δ equals, you can always find an |x| between 0 and δ such that |1/x|>|L|+ε.

When you try to come up with a delta-epsilon proof, typically you work backwards. You assume the epsilon condition holds and find some condition on x from which you can choose an appropriate δ. In this case, you can't find such a δ because the inequality turns out backwards.
 
  • #7
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The strange thing is, the first time through I got the same solution as Spivak, but looking over it again the logic seems downright wrong.

I have those kinds of days too :) lol
 
  • #8
Let me try again: because there are values for delta (e.g. <1/(|L| + E)) that violate the inequality that would result if the limit were true, then the limit is not true by contradiction?
 
  • #9
vela
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No, that's not right. Even when a limit exists, you can usually find a δ for which the ε-inequality won't hold: just use a big δ. The limit doesn't exist because there are no values of δ for which the ε-inequality holds.
 
  • #10
No, that's not right. Even when a limit exists, you can usually find a δ for which the ε-inequality won't hold: just use a big δ. The limit doesn't exist because there are no values of δ for which the ε-inequality holds.

Yeah, I think I figured it out a few hours ago (I was doing another problem and realized what I said here was wrong). It's because we've found a d within which no x will work (all x results in an inconsistency, no matter how small; in fact, the smaller the x the worse the inconsistency becomes)?
 
  • #11
vela
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Almost. You don't need to find a δ such that all values of x within the interval will cause the ε inequality to fail, though in this case you can. You only need one value of x for which inequality fails.
 

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